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Joey throws a dice 3 times and decides to invite

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Joey throws a dice 3 times and decides to invite [#permalink]

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New post 23 Apr 2015, 06:59
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Joey throws a dice 3 times and decides to invite the same number of guests for his birthday party as the sum of the outcomes of the throws. What’s the probability that he invites 12 guests?

    A. 1/6
    B. 1/16
    C. 1/18
    D. 1/36
    E. 25/216
[Reveal] Spoiler: OA

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Re: Joey throws a dice 3 times and decides to invite [#permalink]

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New post 23 Apr 2015, 09:30
Hi,

I tried this one. I am getting a total of 16 combinations to attain 12.

Hence, the answer would be 16/216. But I am missing something somewhere. Please could you explain.

PrepTap wrote:
Joey throws a dice 3 times and decides to invite the same number of guests for his birthday party as the sum of the outcomes of the throws. What’s the probability that he invites 12 guests?

    A. 1/6
    B. 1/16
    C. 1/18
    D. 1/36
    E. 25/216

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Re: Joey throws a dice 3 times and decides to invite [#permalink]

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New post 23 Apr 2015, 09:37
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PrepTap wrote:
Joey throws a dice 3 times and decides to invite the same number of guests for his birthday party as the sum of the outcomes of the throws. What’s the probability that he invites 12 guests?

    A. 1/6
    B. 1/16
    C. 1/18
    D. 1/36
    E. 25/216



Total outcomes:
6^3.

Favorable outcomes:
{6 - 5 - 1} - 6 cases ({6 - 5 - 1}, {6 - 1 - 5}, {5 - 6 - 1}, {5 - 1 - 6}, {1 - 6 - 5}, {1 - 5 - 6});
{6 - 4 - 2} - 6 cases;
{6 - 3 - 3} - 3 cases;
{5 - 5 - 2} - 3 cases;
{5 - 4 - 3} - 6 cases;
{4 - 4 - 4} - 1 case.

6 + 6 + 3 + 3 + 6 + 1 = 25.

P = (favorable)/(total) = 25/6^3.

Answer: E.
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Kudos [?]: 128870 [0], given: 12183

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Re: Joey throws a dice 3 times and decides to invite [#permalink]

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New post 28 Apr 2015, 10:08
I too arrived at the same solution, although it took me close to about 4 minutes to do it. isn't there a faster shortcut for the same?

Bunuel wrote:
PrepTap wrote:
Joey throws a dice 3 times and decides to invite the same number of guests for his birthday party as the sum of the outcomes of the throws. What’s the probability that he invites 12 guests?

    A. 1/6
    B. 1/16
    C. 1/18
    D. 1/36
    E. 25/216



Total outcomes:
6^3.

Favorable outcomes:
{6 - 5 - 1} - 6 cases ({6 - 5 - 1}, {6 - 1 - 5}, {5 - 6 - 1}, {5 - 1 - 6}, {1 - 6 - 5}, {1 - 5 - 6});
{6 - 4 - 2} - 6 cases;
{6 - 3 - 3} - 3 cases;
{5 - 5 - 2} - 3 cases;
{5 - 4 - 3} - 6 cases;
{4 - 4 - 4} - 1 case.

6 + 6 + 3 + 3 + 6 + 1 = 25.

P = (favorable)/(total) = 25/6^3.

Answer: E.

Kudos [?]: [0], given: 35

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Joey throws a dice 3 times and decides to invite [#permalink]

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New post 28 Apr 2015, 11:59
kelvind13 wrote:
I too arrived at the same solution, although it took me close to about 4 minutes to do it. isn't there a faster shortcut for the same?

Bunuel wrote:
PrepTap wrote:
Joey throws a dice 3 times and decides to invite the same number of guests for his birthday party as the sum of the outcomes of the throws. What’s the probability that he invites 12 guests?

    A. 1/6
    B. 1/16
    C. 1/18
    D. 1/36
    E. 25/216



Total outcomes:
6^3.

Favorable outcomes:
{6 - 5 - 1} - 6 cases ({6 - 5 - 1}, {6 - 1 - 5}, {5 - 6 - 1}, {5 - 1 - 6}, {1 - 6 - 5}, {1 - 5 - 6});
{6 - 4 - 2} - 6 cases;
{6 - 3 - 3} - 3 cases;
{5 - 5 - 2} - 3 cases;
{5 - 4 - 3} - 6 cases;
{4 - 4 - 4} - 1 case.

6 + 6 + 3 + 3 + 6 + 1 = 25.

P = (favorable)/(total) = 25/6^3.

Answer: E.


As such there is no shortcut possible but the counting can be done in an orderly fashion which might reduce the time you took.

We know that the sum required is 12.
So if in the first place we have:
1 -> Sum of other two numbers = 12 -1 = 11 -> possible combinations (5+6, 6+5)
2 -> Sum of other two numbers = 12 -2 = 10 -> possible combinations (4+6, 6+4, 5+5)
3 -> Sum of other two numbers = 12 -3 = 9 -> possible combinations (3+6, 6+3, 4+5, 5+4)
4 -> Sum of other two numbers = 12 - 4 = 8 -> possible combinations (2+6, 6+2, 3+5, 5+3, 4+4)
5 -> Sum of other two numbers = 12 -5 = 7 -> possible combinations (1+6, 6+1, 2+5, 5+2, 3+4, 4+3)
6 -> Sum of other two numbers = 12 -6 = 6 -> possible combinations (1+5, 5+1, 2+4, 4+2, 3+3)

Favourable outcomes: 25
Total outcomes: 6 * 6 * 6 = 216

Probability = 25/216

Last edited by PrepTap on 07 May 2015, 00:04, edited 1 time in total.

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Re: Joey throws a dice 3 times and decides to invite [#permalink]

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New post 28 Apr 2015, 17:03
Min sum=1+1+1=3
Max sum=6+6+6=18
clearly there is only 1 way to achieve these sums.

Sum=4; #ways 3
sum=17; #ways 3

So we know that the series of #ways for sum {3..10} will be just reverse of series of #ways for sum {11..18}
and each series sums to 108.

sum=9 or 12 #ways will be same.

Sum=9 ; lets look at it.

A+b+c=9
Now this is just a question of distributing 9 identical things into 3 people. Since each one will have atleast 1 thing so we have to distribute 6 things among 3 people.
8C3=28
now there will be 3 cases in which each a, b and c will have 7 things so subtract 3 from 28 To get 25.

25/216 answer
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Kudos [?]: 538 [0], given: 75

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Joined: 02 Jan 2017
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Re: Joey throws a dice 3 times and decides to invite [#permalink]

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New post 03 Oct 2017, 16:34
kelvind13 wrote:
I too arrived at the same solution, although it took me close to about 4 minutes to do it. isn't there a faster shortcut for the same?

Bunuel wrote:
PrepTap wrote:
Joey throws a dice 3 times and decides to invite the same number of guests for his birthday party as the sum of the outcomes of the throws. What’s the probability that he invites 12 guests?

    A. 1/6
    B. 1/16
    C. 1/18
    D. 1/36
    E. 25/216



Total outcomes:
6^3.

Favorable outcomes:
{6 - 5 - 1} - 6 cases ({6 - 5 - 1}, {6 - 1 - 5}, {5 - 6 - 1}, {5 - 1 - 6}, {1 - 6 - 5}, {1 - 5 - 6});
{6 - 4 - 2} - 6 cases;
{6 - 3 - 3} - 3 cases;
{5 - 5 - 2} - 3 cases;
{5 - 4 - 3} - 6 cases;
{4 - 4 - 4} - 1 case.

6 + 6 + 3 + 3 + 6 + 1 = 25.

P = (favorable)/(total) = 25/6^3.

Answer: E.



I solved it in 2 mins 10 secs using this approach. hope it helps.

You will have to involve bit of permutations for a shortcut, assume each setting of 3 throws as letters. e.g 444 is a letter repeating thrice 255 is a letter with two repeating elements.
Having that in mind.
List out basic settings which give you and permute for repeating elements

(1,5,6) This can be written in 3 ! ways [ as it has no repeating elements] 6 ways
(2,5,5) this can be written in 3 !/2! ways [ as it has two repeating elements] 3 ways
(2,4,6) This can be written 3 ! ways. 6 ways
(3,4,5) Same 3 ! ways 6 ways
(444) This can be written in 3!/3! ways 1 way
(6,3,3) This can be written in 3!/2! ways 3 ways


Total outcomes = 6^3 = 216
Possible outcomes = 6+6+6+3+3+1= 25
Answer is E 25/216

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Re: Joey throws a dice 3 times and decides to invite   [#permalink] 03 Oct 2017, 16:34
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